c program unsigned int or uint8_t

Question

I am writing a function called take_picture() . The function will make a camera take a picture and store the data in an array.

I want the function to return the address where the array is located and to return the size of the array.

So this is my function declaration:

uint8_t * take_picture(int *piclength)

My question is what would be the correct/logical data types for the function return type and the argument? considering that they store an address and size of an array? Does a uint8_t and int make sense?

Thanks

Answer 1

My question is what would be the correct/logical data types for the function return type and the argument? considering that they store an address and size of an array?

The function return type should be a pointer to an array item. If it is an array of uint8_t , then the return type should be uint8_t* . This is also the most correct/safe type for an array of bytes ("raw data").

The most correct type used to describe the size of an array is size_t , an unsigned integer type that exists specifically for this purpose, and is guaranteed portably to be large enough to hold an array size of the given system.

uint8_t* take_picture (size_t* piclength);

or alternatively

void take_picture (uint8_t* picture, size_t* piclength);

Answer 2

unsigned int size; // As the image size is always positive and is large.

uintptr_t address; // if this is available in stdint.h

or

unsigned int address; //it is capable of storing a pointer size.

address = take_picture(&size); //take_picture if returns uintptr_t

If you do not know what take_picture returns, then you cast it like what you have done:

address = (unsigned *int)take_picture(& size);